(x^2)+(8x)=0

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Solution for (x^2)+(8x)=0 equation: 



(x^2)+(8x)=0

a = 1; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·1·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*1}=\frac{-16}{2}
=-8
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*1}=\frac{0}{2}
=0
$

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